Two point charges q1 = 2 μC and q2 = -3 μC are separated by r = 0.05 m. What is the electric potential energy of the system? (Use k = 8.99×10^9 N·m^2/C^2.)

Two point charges q1 = 2 μC and q2 = -3 μC are separated by r = 0.05 m. What is the electric potential energy of the system? (Use k = 8.99×10^9 N·m^2/C^2.)

• A. -1.08 J
• B. -0.54 J
• C. +1.08 J
• D. -2.16 J

Answer: (A)

The quantity tested is how electric potential energy between two point charges depends on their charges and their separation. For two point charges, the potential energy is U = k q1 q2 / r. The sign comes from the product q1 q2: opposite charges give a negative product, so the energy is negative, reflecting the attraction between them. Plug in the numbers: q1 = 2 μC = 2×10^-6 C, q2 = -3 μC = -3×10^-6 C, r = 0.05 m, k = 8.99×10^9 N·m^2/C^2. q1 q2 = (2×10^-6)(-3×10^-6) = -6×10^-12 C^2 k q1 q2 = 8.99×10^9 × (-6×10^-12) ≈ -0.05394 J·m U = (k q1 q2) / r ≈ (-0.05394) / 0.05 ≈ -1.0788 J ≈ -1.08 J So the system’s electric potential energy is about -1.08 joules. The negative sign confirms attraction between the opposite charges.

The quantity tested is how electric potential energy between two point charges depends on their charges and their separation. For two point charges, the potential energy is U = k q1 q2 / r. The sign comes from the product q1 q2: opposite charges give a negative product, so the energy is negative, reflecting the attraction between them.

Plug in the numbers: q1 = 2 μC = 2×10^-6 C, q2 = -3 μC = -3×10^-6 C, r = 0.05 m, k = 8.99×10^9 N·m^2/C^2.

q1 q2 = (2×10^-6)(-3×10^-6) = -6×10^-12 C^2

k q1 q2 = 8.99×10^9 × (-6×10^-12) ≈ -0.05394 J·m

U = (k q1 q2) / r ≈ (-0.05394) / 0.05 ≈ -1.0788 J ≈ -1.08 J

So the system’s electric potential energy is about -1.08 joules. The negative sign confirms attraction between the opposite charges.